An electric dipole is placed as shown in the figure.The electric potential (in $10^2\,V$ ) at point $P$ due to the dipole is $\left(\epsilon_0=\right.$ permittivity of free space and $\left.\frac{1}{4 \pi \epsilon_0}=K\right)$ :
NEET 2023, Medium
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$v =\frac{ Kq }{2 \times 10^{-2}}-\frac{ Kq }{8 \times 10^{-2}}$
$= Kq \left[\frac{3}{8}\right] \times 10^2$
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